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x+log(x)=y−1y+log(y−1)=z−1z+log(z−2)=x+2 \begin{aligned} x+\log{(x)} = y-1\\ y+\log{(y-1)} = z-1 \\ z+\log{(z-2)} = x+2 \\ \end{aligned}x+log(x)=y−1y+log(y−1)=z−1z+log(z−2)=x+2
There is only one solution set (x,y,z)\left(x,y,z\right)(x,y,z) to the system of equations above. Evaluate x+y+zx+y+zx+y+z.
Note: Take the base of the logarithms as 10.
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